Let the function f ( x ) = 2 x 3 + ( 2 p - 7 ) x 2 + 3 ( 2 p - 9 ) x - 6 have a maxima for some value of x 0 and a minima for some value of x 0. Then, the set of all values of p is [2023]

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Q.
Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$ have a maxima for some value of $x$ < 0 and a minima for some value of $x$ > 0. Then, the set of all values of $p$ is [2023]

1 $(\frac{9}{2}, \infty)$

2 $(0, \frac{9}{2})$

3 $(- \frac{9}{2}, \frac{9}{2})$

4 $(- \infty, \frac{9}{2})$

Ans.
(4)

We have, $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$

$Differentiating w.r.t. x, we get$

$f' (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$

$∵$ $f' (0) < 0 \Rightarrow 3 (2 p - 9) < 0$

$\Rightarrow p < \frac{9}{2}$

$So, p \in (- \infty, \frac{9}{2})$

Solution

Q.
Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$ have a maxima for some value of $x$ < 0 and a minima for some value of $x$ > 0. Then, the set of all values of $p$ is [2023]

1 $(\frac{9}{2}, \infty)$

2 $(0, \frac{9}{2})$

3 $(- \frac{9}{2}, \frac{9}{2})$

4 $(- \infty, \frac{9}{2})$

Ans.
(4)

We have, $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$

$Differentiating w.r.t. x, we get$

$f' (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$

$∵$ $f' (0) < 0 \Rightarrow 3 (2 p - 9) < 0$

$\Rightarrow p < \frac{9}{2}$

$So, p \in (- \infty, \frac{9}{2})$

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Solution

Q. Let the function f(x)=2x3+(2p-7)x2+3(2p-9)x-6 have a maxima for some value of x < 0 and a minima for some value of x > 0. Then, the set of all values of p is [2023]

1 (92,∞)

2 (0,92)

3 (-92,92)

4 (-∞,92)

Ans. (4) We have, f(x)=2x3+(2p-7)x2+3(2p-9)x-6 Differentiating w.r.t. x, we get f'(x)=6x2+2(2p-7)x+3(2p-9) ∵ f'(0)<0⇒3(2p-9)<0 ⇒p<92 So, p∈(-∞,92)

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Q.
Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$ have a maxima for some value of $x$ < 0 and a minima for some value of $x$ > 0. Then, the set of all values of $p$ is [2023]

1 $(\frac{9}{2}, \infty)$

2 $(0, \frac{9}{2})$

3 $(- \frac{9}{2}, \frac{9}{2})$

4 $(- \infty, \frac{9}{2})$

Ans.
(4)

We have, $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$

$Differentiating w.r.t. x, we get$

$f' (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$

$∵$ $f' (0) < 0 \Rightarrow 3 (2 p - 9) < 0$

$\Rightarrow p < \frac{9}{2}$

$So, p \in (- \infty, \frac{9}{2})$